Let complex numbers α and 1 α ¯ lie on circles x - x 0 2 + y - y 0 2 = r 2 ,  and x - x…

Let complex numbers α and 1 α ¯   lie on circles x - x 0 2 + y - y 0 2 = r 2 ,   and   x - x 0 2 + y - y 0 2 = 4 r 2 , respectively. If z 0 = x 0 + i⁡ y 0  satisfies the equation 2 z 0 2 = r⁡ 2 + 2 , then α =
  1. 1 2
  2. 1 2
  3. 1 7
  4. 1 3

Solution

z - z 0 = r
z - z 0 = 2 r
α - z 0 = r
1α¯-z0=2rαα¯=α2
α α 2 - z 0 = 2 r
α - z 0 α ¯ - z ¯ 0 = r 2 ⇒ α 2 - z 0 α ¯ - α z ¯ 0 + z 0 2 = r 2
α α 2 - z 0 α ¯ α 2 - z ¯ 0 = 4 r 2 ⇒ α 2 α 4 - z ¯ 0 α α 2 - z ¯ 0 α α 2 + z 0 2 = 4 r 2
1 - z 0 α ¯ - z ¯ 0 α + z 0 2 α 2 = 4 r 2 α 2
⇒ α 2 - 1 + z 0 2 1 - α 2 = r 2 1 - 4 α 2
α 2 - 1 1 - r 2 + 2 2 = r 2 1 - 4 α 2
α 2 - 1 - r 2 2 = r 2 1 - 4 α 2
α 2 - 1 = - 2 + 8 α 2
1 = 7 α 2 ⇒ α = 1 7

Asked in: JEE Advanced 2013 (Paper 1)

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